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The problem asks to use the limit property:

\[
\lim_{x \to 0} \frac{\sin x}{x} = 1
\]

to show that:

\[
\lim_{bx \to 0} \frac{\sin(bx)}{bx} = 1.
\]

### Step-by-Step Solution:

1. **Recognize the Given Formula**:
   - The given limit property is \(\lim_{x \to 0} \frac{\sin x}{x} = 1\). This is a well-known trigonometric limit.

2. **Rewrite the Expression**:
   - The goal is to show that \(\lim_{bx \to 0} \frac{\sin(bx)}{bx} = 1\).
   - Here, \(bx\) behaves like a single variable approaching zero. Let \(u = bx\). As \(x \to 0\), \(u \to 0\) as well.

3. **Substitute and Simplify**:
   - Rewrite the limit expression in terms of \(u\):
   \[
   \lim_{bx \to 0} \frac{\sin(bx)}{bx} = \lim_{u \to 0} \frac{\sin(u)}{u}.
   \]
   - Since this is in the exact form of the original limit property, we can directly apply it.

4. **Apply the Limit Property**:
   - By the known limit:
   \[
   \lim_{u \to 0} \frac{\sin(u)}{u} = 1.
   \]
   - Therefore:
   \[
   \lim_{bx \to 0} \frac{\sin(bx)}{bx} = 1.
   \]

This completes the proof that \(\lim_{bx \to 0} \frac{\sin(bx)}{bx} = 1\).

---

Would you like further details or have any questions?

Here are 5 related questions to deepen your understanding:

1. What other trigonometric limits are important in calculus?
2. How does L'Hôpital's Rule apply to solving limits like this one?
3. What happens to the limit if \(b\) is negative?
4. How can this limit be used to find the derivative of \(\sin(x)\)?
5. What is the geometric interpretation of this limit?

**Tip:** When proving limits involving sine, always check if the limit can be transformed into the standard form \(\frac{\sin x}{x}\) as \(x \to 0\).

Use \( \lim_{{x \to 0}} \frac{{\sin x}}{x} = 1 \) to show that \( \lim_{{bx \to 0}} \frac{{\sin(bx)}}{bx} = 1 \).

Learn how to prove the limit \( \lim_{{bx \to 0}} \frac{{\sin(bx)}}{bx} = 1 \) by using the fundamental limit \( \lim_{{x \to 0}} \frac{{\sin x}}{x} = 1 \). This solution covers the step-by-step approach to simplifying and applying known trigonometric limits, ideal for students in Grades 11-12.

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